Question

6 sin²x - 3 sin x*cos x - 5 cos²=2

Answer 1

6sin²x-3sinxcosx-5cos²x=2;

6sin²x-3sinxcosx-5cos²x-2sin²x-2cos²x=0;

4sin²x-3sinxcosx-7cos²x=0|:cos²x {cos²x≠0};

4tg²x-3tgx-7=0;

D=9+112=121; √D=11;

tgx=(3±11)/8=1,75; -1.

x=arctg(1.75)+πn. n∈Z.

x=-π/4+πn. n∈Z.

Answer 2

6 sin²x - 3 sin x*cos x - 5 cos²=2

  6 sin²x - 3 sin x*cos x - 5 cos²-2=0 

 6sin²x-3sinxcosx-5cos²x-2sin²x-2cos²x=0

 4sin²x-3sinxcosx-7cos²x=0 /:cos²x

 4tg²x-3tgx-7=0

пусть tgx=t

 4t² - 3t - 7 = 0
D = b²- 4ac
D = 9 + 112 = 121 = 11²

t1,2 = -b ± √D/2a
t1 = 3 + 11/8 = 14/8 = 7/4
t2 = 3 - 1/8 = - 8/8= -1
Ответ: x=arctg(7/4) пиn

           x=-пи/4+пиn